Theory

Improper integrals

How the integral extends to unbounded intervals and to unbounded functions, which benchmark integrals serve as reference, and how the comparison tests establish convergence without computing an antiderivative.

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Integral over an unbounded interval
Integral of an unbounded function

Integral over an unbounded interval

Let $f : [a, + \infty) \to R$ be integrable on $[a, t]$ for every $t > a$ . The improper integral of $f$ over $[a, + \infty)$ is

\int_{a}^{+ \infty} f (x) d x = t \to + \infty lim \int_{a}^{t} f (x) d x .

The integral is called

convergent, if the limit exists and is finite;
divergent, if the limit is $+ \infty$ or $- \infty$ ;
indeterminate, if the limit does not exist.

On $(- \infty, b]$ the definition uses the limit as $t \to - \infty$ of the integral over $[t, b]$ . Establishing which of the three cases an integral falls into, that is, its behavior, is the problem of this chapter.

y = \dfrac{1}{x^{2}}

The integral over

[1, t]

is the shaded area under

y = \frac{1}{x ^{2}}

and equals

1 - \frac{1}{t}

; the improper integral is the limit of this area as

t \to + \infty

, that is,

1

The three behaviors

\int_{1}^{t} \frac{d x}{x ^{2}} = 1 - \frac{1}{t} \to 1, \int_{1}^{t} \frac{d x}{x} = ln t \to + \infty, \int_{0}^{t} cos x d x = sin t .

The first integral converges and equals $1$ ; the second diverges; the third is indeterminate, because $sin t$ as $t \to + \infty$ oscillates between $- 1$ and $1$ without admitting a limit.

Remark

The behavior depends only on what happens near the critical point. For $a < b$ the integrals over $[a, + \infty)$ and over $[b, + \infty)$ differ by the ordinary integral over $[a, b]$ , therefore they have the same behavior. For this reason the tests in this chapter require their hypotheses only eventually, that is, from some point on.

Integral of an unbounded function

Integral of a function unbounded near an endpoint

Let $f : (a, b] \to R$ be integrable on $[t, b]$ for every $t \in (a, b]$ and unbounded near $a$ . One sets

\int_{a}^{b} f (x) d x = t \to a^{+} lim \int_{t}^{b} f (x) d x,

with the same terminology: convergent, divergent, indeterminate. If $f$ is unbounded near $b$ , one uses the limit as $t \to b^{-}$ of the integral over $[a, t]$ . An endpoint near which the function is unbounded, as well as an infinite endpoint, is called a critical point of the integral.

Example

\int_{t}^{1} \frac{d x}{x} = - ln t \to + \infty (t \to 0^{+}), \int_{t}^{1} \frac{d x}{x} = 2 - 2 t \to 2.

The integral of $\frac{1}{x}$ over $(0, 1]$ diverges, while that of $\frac{1}{x}$ converges and equals $2$ , even though both integrands tend to $+ \infty$ as $x \to 0^{+}$ .

y = \dfrac{1}{\sqrt{x}}

y = \dfrac{1}{x}

(0, 1]

the shaded area under

y = \frac{1}{x}

is finite and equals

2

; under

y = \frac{1}{x}

(dashed), which grows faster near

0

, the area is infinite

The theory continues with 7 more sections (several critical points, benchmark integrals, comparison test…), plus all worked examples, the solutions and the PDF.

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